Find the vector $\mathbf { v }$ that has a magnitude of 8 and is in the same direction as $\mathbf { u }$, where $\mathbf { u } = \langle - 6 , - 4 \rangle$.
A) $\mathbf { v } = \left\langle \frac { - 24 } { \sqrt { 13 } } , \frac { - 16 } { \sqrt { 13 } } \right\rangle$
B) $\mathbf { v } = \left\langle \frac { - 3 } { \sqrt { 13 } } , \frac { - 2 } { \sqrt { 13 } } \right\rangle$
C) $\mathbf { v } = \left\langle \frac { - 8 } { \sqrt { 13 } } , \frac { - 4 } { \sqrt { 13 } } \right\rangle$
D) $\mathbf { v } = \left\langle \frac { - 2 } { \sqrt { 13 } } , \frac { 2 } { \sqrt { 13 } } \right\rangle$
E) $\mathbf { v } = \left\langle \frac { 4 } { \sqrt { 13 } } , \frac { 6 } { \sqrt { 13 } } \right\rangle$

Question

Quizplus · Accepted Answer

To find a vector with a specific magnitude in the same direction as another vector, you first normalize the original vector and then multiply by the desired magnitude. The magnitude of $\mathbf{u}$ is $\sqrt{(-6)^2 + (-4)^2} = \sqrt{36 + 16} = \sqrt{52} = \sqrt{4 \cdot 13} = 2\sqrt{13}$. Normalizing $\mathbf{u}$ gives $\left\langle \frac{-6}{2\sqrt{13}}, \frac{-4}{2\sqrt{13}} \right\rangle = \left\langle \frac{-3}{\sqrt{13}}, \frac{-2}{\sqrt{13}} \right\rangle$. Multiplying this by 8 (to get the desired magnitude) gives $\left\langle \frac{-24}{\sqrt{13}}, \frac{-16}{\sqrt{13}} \right\rangle$.

Find the Vector v\mathbf { v }v That Has a Magnitude of 8 and Is in the Magnitude

Find the Vector $\mathbf { v }$ That Has a Magnitude of 8 and Is in the Magnitude